221

= 2(.153)(152.5) = 46.6 kips. Thus, the vertical component Vp is

46.6*sin(2.44) = 1.98 kips. Showr. in Table 4.15 are the values of the

additional contribution to the shear strength of the member due to the

presence of draped strands Vp.

For those regions of the member in the uncracked or transition

state, where the concrete in the web provides addi tional shear strength

Vc ' Eq. 4.43 becomes

(4.46)

Rearranging Eq. 4.44 results in

(IL 47)

Since ns = zcota and Sy = Avfy

(11.48)

where Avis is the area of stirrups resisting the factored shear force

per inch of the stirrup spacing "s", and f y is the yield strength of the

stirrup reinforcement. For this design example, fy = 60000 psi, a=

41.8 degrees, hence tan a = 0.90.

Eq. 4.48 is used to design the web reinforcement required to

resist the factored shear force. Shown in row (7) of Table 4.16 are the

required amounts of web reinforcement per inch of stirrup spacing "s"

for each of the design zones. Row (8) contains the minimum amount of

web reinforcement which must be provided whenever the factored shear

stress (V u) exceeds the value 1.0¢/f[, where ¢ = 0.85. The minimum

222

amount of web reinforcement is evaluated in accordance with the

requirements of Sec. 1.4.2.1 of tne proposed design recommendations.

Hence,

(4.49)

As can be seen from Table 4.15, the value of the shear stress due to the

factored shear force (vu[V]) at all the design sections exceeds the

value of 1.0 ¢./fb = 1.0(0.85) ./5000 = 0.06 ksi, hence at least the

minimum amount of web reinforcement must be provided in all the design

zones.

Row (9) of Table 4.16 shows the required stirrup spacing if a

Grade 60 113 U stirrup is used as web reinforcement. Row (10) indicates

the maximum allowed stirrup spacing as required by Sec. 1.4.2.6 of the

proposed design recommendations. Therefore, in the design zone 1-2 113 U

stirrups at 9.5 inches center-to-center should be provided. In the

design zones 2-3, 3-4, 4-5, and 5-6, 113 U's at 12 inches must be

provided. The U stirrups shall be terminated in the compression zone

with a 135 degree hook at the ends.

4.4.4 Evaluation of the Compression Stresses in the Fan

Regions. As explained in Secs. 4.3.5, the presence of concentrated load

disturbs the continuous uniform compression field of the truss. The

presence of a concentrated load introduces a series of diagonal

compression forces which fan out from the concentrated load. Hence, in

this design example compression fans will form at both supports where

the reaction introduces compression, and under the concentrated truck

223

wheel loads. As previously explained, the geometry of the compression

fan depends on the spacing of the transverse reinforcement and the

chosen angle 0'. Figure 4.45 shows the compression "fan" generated at

the supports of the composi te I gird er. Col urn n (5) of Table 4.17 sho ws

the compression forces generated at each of the joints of the truss in

the compression fan zone. Column (6) shows the diagonal compression

(1) (2) (3) (4) (5) (6)

Point a (1) tan a (1) Di =

Si [S1 + wns] fdi (1) sin a (i)

( degrees) (kips) (ksi)

83.7 9.1 Sy 14.79 0.450

2 71.4 2.97 Sy 15.51 0.160

3 61 • 1 1.81 Sy 16.79 0.120

4 52.3 1.29 Sy 18.58 0.10

5 45.2 1.01 Sy 20.72 0.10

Table 4.17 Diagonal compression stresses in the fan region

stresses induced by the diagonal compression forces shown in column (5).

As previously explained, the diagonal compression stresses at each of

the joints (i) of the truss is given as:

(4.50 )

224

Z=43"

4.75"9.5" 9.5" 9.5" 9.5" I '" ~-C ~~ ~4 .-1-4 ~k .14

5=9..5" nS

48" ~I

Fig. 4.45 Compression fan at the support

v

s

~F L